With that formula, speed could never be zero, obviously. It seems the cutoff Fozzie uses is around 100 m/s.

You're mostly correct, yes. The exact time depends on the max (sub-warp) speed of the ship. Obviously this varies a lot even within a single ship class, and even more so depending on fittings. Ships will exit warp mode when their warping speed drops below 50% of sub-warp max speed, or 100m/s, whichever is the lower.
My chart is written assuming a simplified fixed exit at 100m/s, so it sounds like your math is good :)

IIRC 100m/s or 50% of sub-warp speed. I lost my bookmarks, so let me see if I can dig up the reference.



reference: https://forums.eveonline.com/default.aspx?g=posts&m=3902148#post3902148

No, we are talking about the moment the ship enters the warp.

OP, Fozzie's formulas are fine.

x = e^kt is distance, in meters; t is time in seconds

v = k*e^kt is speed, in m/s; this formula is the derivative of the above formula, aka the definition of speed. it's correct.


Now, say Vmax is your maximum warp speed. Vmax = k*A where A is 150 billion meters (1 AU).

So if you want to know acceleration time, its Vmax = k*A = k*e^kt.

==> A = e^kt

==> ln(A) = kt

==> t = ln(A)/k = 25.7 / k


For example, a cruiser (3 AU/s) would take 8.6 seconds to reach max warp speed, an inty (8 AU/s) would take 3.2 seconds.

Sounds about right to me, though I CBA to test it.


Fozzie was slightly fuzzier on deceleration, I'll think about it and post back if I nail the formula.

Ok, that was my initial reaction. Looking further into it, you may be right about it. Even disregarding boundary conditions, those formulas are wrong dimensions-wise. e is dimensionless, while x is meters, so x=e^anypower is wrong. the power of e should be dimensionless as well, but kt has distance dimensions, etc.

Anyway, in this light, Vmax as kA may be correct as it converts AU to meters. Then taking into account the corrected formula we do get warp acceleration time as
t_warp=ln(A+1)/k
and it sounds about right for my rigged badger with 6 point something warp speed.
So that part you got right. The distance though to reach max speed would be different.
11 seconds for deceleration though(same as interceptor, since k for deceleration is min(2,warp_speed/3)? I will check in game later today. It seems way to long. And it always looked to me like when the ship is out of warp, the speed gauge is at about 1/4. Could be different for different ships though, I will check in game later today!

Well, thank you Gully for staying on topic and actually contributing, unlike 20+ other responses in this thread!

Yw!

Tbh, I was about to post a useless but witty reply too, but then I noticed the server issues/extended downtime and decided it would be more fun to try to work out the maths.

The formulas I've posted give exactly the same figures of Fozzie's spreadsheet, so they should be quite ok.

Yes, it takes inties 11 seconds to decelerate. The deceleration constant is capped at 2 no matter the max warp speed - even if, say, you put 3 T2 warp speed rigs on an inty. It's pretty clear in the second graph here.

I just noticed j=k/3 up to k=6, so that means deceleration distance is exactly 3 AU up to 6AU/s max speed.

I previously thought that j (decel constant) was fixed for each hull type. I edited my previous post accordingly.


This also means that fitting 3x warp speed rigs to a cruiser actually does make it warp almost as fast as a frigate - interesting.